Chemistry (Math in Disguise of Science)

Flashcard maker : Lily Taylor
Calculate the mass in grams 1.25 moles of aluminum chloride.
Calculate the mass in grams 1.25 moles of aluminum chloride.
1. Add the atomic mass, multiply them if they have a little number.
2. Multiply the number they gave you over 1 GRAM and the number you got from adding the masses over 1 MOLE.
EX
Al-26.98
Cl-(3)35.45
133.33

1.25 mol/1 g X 133.33 g/1 mol= 166.62 g /mol

Calculate the number of molecules present in 4.75 g of phosphine , PH3(little 3).
Calculate the number of molecules present in 4.75 g of phosphine , PH3(little 3).
1. Add the atomic mass, multiply them if they have a little number.
2. Multiply the number they gave you over 1 MOLECULE and the AVAGOTES number over the number you got.

EX
P-30.97
H-(3)1.0079
33.99 g

4.75g/1 Molecules X 6.22×10^23/33.99 g = 8.42×10^23 Molecules

Calculate the percent by mass of Methane CH4(little 4).
Calculate the percent by mass of Methane CH4(little 4).
1. Add the atomic mass, multiply them if they have a little number.
2. Atomic mass g over the number you got.
3. Times 100

EX
C-12.01
H-(4)1.0079
16.04 g

C- 12.01gC/16.04g X 100= 75%
H- 4.03gH/16.04g X 100=25%

A compound was analyzed and found to contain the following .
Baruim-89.59%
Oxygen-10.44%
Determine the empirical formula of it.
A compound was analyzed and found to contain the following .
Baruim-89.59%
Oxygen-10.44%
Determine the empirical formula of it.
1. Change the percent to grams.
2. Put the number they gave you over 1 MOLE and multiply it by 1 MOLE over its atomic mass.
3. Put the number you got over the smallest number.
4. The number you now have is the subscript.

EX
Ba-89.59%=89.59g
O-10.44%=10.44g

89.59g Ba/ 1 mol X 1 mol/ 137.3 g = 0.6253 mol Ba
10.44g O/ 1 mol X 1 mol/16.00 g = 0.6253 mol O

0.6253 mol Ba/0.6253 mol =1 mol Ba
0.6253 mol O/0.6253 mol = 1 mol O
=
BaO

An oxide of aluminum if formed by the reaction of 4.15 g of aluminum with 3.692 g of oxygen. Calculate the empirical formula for this compound.
An oxide of aluminum if formed by the reaction of 4.15 g of aluminum with 3.692 g of oxygen. Calculate the empirical formula for this compound.
1. Put the number they gave you over 1 mole times 1 mol over the atomic mass.
2. Divide both by the smallest number.
3. If you don’t get a full number , than multiply ALL till you do.

EX
Al- 4.151g Al/1 mol X 1 mol/26.98 g Al = .1538
O- 3.692g O/1 mol X 1 mol/ 16 G O= .23075

.1538/ .1538 = 1 X 2= 2
.23075/.1538=1.5 X 2=3
=Al2O3 ( Little 2 & Little 3)

A compound with the empirical formula CH4O (little 4) was formed in a subsequent experiment to have a molar mass of apprimatldy 192 g. What is the molecule formula of the compound?
A compound with the empirical formula CH4O (little 4) was formed in a subsequent experiment to have a molar mass of apprimatldy 192 g. What is the molecule formula of the compound?
1. Add the atomic mass, multiply them if they have a little number.
2. Put the number they gave you over the number you got.
3. If you got number that info full multiply it till you do.

EX
C-12.01
H-(4)1.008
O-15.9994
32.0414

192g/32.0414g = 5.99= 6
C6H24O6 (little 4, little 24, little 6)

Calculate the number of moles in 6.22 g of Aluminum.
Calculate the number of moles in 6.22 g of Aluminum.
1. Put the number they gave you over 1 mole and times it by 1 mole over the atomic mass

EX
6.22 g Al/ 1 mol X 1 mol/ 26.98 g Al = 0.23 g/mol Al

Calculate the molar mass for magnesium nitride, Mg(NO2)2 ( little 2 ,little 2).
Calculate the molar mass for magnesium nitride, Mg(NO2)2 ( little 2 ,little 2).
1. Add the atomic mass, multiply them if they have a little number.

EX
Mg-24.305
N-(2)14.00674
O-(4)15.9994
116.31

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